Mixing
What is her real age? [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
asked Aug 29 at 9:26
rsp
1,44531026
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
 |Â
show 1 more comment
up vote
4
down vote
favorite
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
asked Aug 29 at 9:26
rsp
1,44531026
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
3
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
1
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
asked Aug 29 at 9:26
rsp
1,44531026
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
asked Aug 29 at 9:26
rsp
1,44531026
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
edited Aug 29 at 10:09
JonMark Perry
13.8k42768
13.8k42768
asked Aug 29 at 9:26
rsp
1,44531026
asked Aug 29 at 9:26
rsp
1,44531026
asked Aug 29 at 9:26
rsp
1,44531026
1,44531026
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
3
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
1
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34
 |Â
show 1 more comment
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
3
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
1
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
3
3
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
1
1
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34
 |Â
show 1 more comment
5 Answers
5
active
oldest
votes
up vote
15
down vote
accepted
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 at 9:41
sedrick
1,656514
add a comment |Â
up vote
8
down vote
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 at 11:43
Bass
22.2k355143
add a comment |Â
up vote
6
down vote
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 at 13:24
Cashbee
65512
add a comment |Â
up vote
4
down vote
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
answered Aug 29 at 14:57
FrazzleUK
1413
add a comment |Â
up vote
1
down vote
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 at 11:41
user477343
3,0601742
answered Aug 29 at 10:08
James
24127
Always make sure to hide your answer(s) in spoiler quotes/tags>!as opposed to>, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
– user477343
Aug 29 at 11:38
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 at 9:41
sedrick
1,656514
add a comment |Â
up vote
15
down vote
accepted
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 at 9:41
sedrick
1,656514
add a comment |Â
up vote
15
down vote
accepted
up vote
15
down vote
accepted
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 at 9:41
sedrick
1,656514
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 at 9:41
sedrick
1,656514
answered Aug 29 at 9:41
sedrick
1,656514
answered Aug 29 at 9:41
sedrick
1,656514
answered Aug 29 at 9:41
sedrick
1,656514
1,656514
add a comment |Â
add a comment |Â
up vote
8
down vote
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 at 11:43
Bass
22.2k355143
add a comment |Â
up vote
8
down vote
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 at 11:43
Bass
22.2k355143
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 at 11:43
Bass
22.2k355143
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 at 11:43
Bass
22.2k355143
answered Aug 29 at 11:43
Bass
22.2k355143
answered Aug 29 at 11:43
Bass
22.2k355143
answered Aug 29 at 11:43
Bass
22.2k355143
22.2k355143
add a comment |Â
add a comment |Â
up vote
6
down vote
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 at 13:24
Cashbee
65512
add a comment |Â
up vote
6
down vote
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 at 13:24
Cashbee
65512
add a comment |Â
up vote
6
down vote
up vote
6
down vote
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 at 13:24
Cashbee
65512
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 at 13:24
Cashbee
65512
answered Aug 29 at 13:24
Cashbee
65512
answered Aug 29 at 13:24
Cashbee
65512
answered Aug 29 at 13:24
Cashbee
65512
65512
add a comment |Â
add a comment |Â
up vote
4
down vote
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
answered Aug 29 at 14:57
FrazzleUK
1413
add a comment |Â
up vote
4
down vote
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
answered Aug 29 at 14:57
FrazzleUK
1413
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
answered Aug 29 at 14:57
FrazzleUK
1413
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
answered Aug 29 at 14:57
FrazzleUK
1413
edited Aug 29 at 15:18
answered Aug 29 at 14:57
FrazzleUK
1413
answered Aug 29 at 14:57
FrazzleUK
1413
answered Aug 29 at 14:57
FrazzleUK
1413
1413
add a comment |Â
add a comment |Â
up vote
1
down vote
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 at 11:41
user477343
3,0601742
answered Aug 29 at 10:08
James
24127
Always make sure to hide your answer(s) in spoiler quotes/tags>!as opposed to>, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
– user477343
Aug 29 at 11:38
add a comment |Â
up vote
1
down vote
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 at 11:41
user477343
3,0601742
answered Aug 29 at 10:08
James
24127
Always make sure to hide your answer(s) in spoiler quotes/tags>!as opposed to>, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
– user477343
Aug 29 at 11:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 at 11:41
user477343
3,0601742
answered Aug 29 at 10:08
James
24127
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 at 11:41
user477343
3,0601742
answered Aug 29 at 10:08
James
24127
edited Aug 29 at 11:41
user477343
3,0601742
edited Aug 29 at 11:41
user477343
3,0601742
edited Aug 29 at 11:41
user477343
3,0601742
3,0601742
answered Aug 29 at 10:08
James
24127
answered Aug 29 at 10:08
James
24127
answered Aug 29 at 10:08
James
24127
24127
Always make sure to hide your answer(s) in spoiler quotes/tags>!as opposed to>, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
– user477343
Aug 29 at 11:38
add a comment |Â
Always make sure to hide your answer(s) in spoiler quotes/tags>!as opposed to>, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
– user477343
Aug 29 at 11:38
Always make sure to hide your answer(s) in spoiler quotes/tags
>! as opposed to >, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)– user477343
Aug 29 at 11:38
Always make sure to hide your answer(s) in spoiler quotes/tags
>! as opposed to >, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)– user477343
Aug 29 at 11:38
add a comment |Â
Today 35 years old, correct?
– Doomenik
Aug 29 at 9:31
@Doomenik Yes...
– rsp
Aug 29 at 9:32
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
– rhsquared
Aug 29 at 9:40
3
How is this a puzzle and not merely an arithmetic story-problem?
– Rubio♦
Aug 29 at 13:32
1
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
– Bass
Aug 29 at 15:34